A curve passes through the point ($$x$$ = 1, $$y$$ = 0) and satisfies the differential equation

$${{dy} \over {dx}} = {{{x^2} + {y^2}} \over {2y}} + {y \over x}$$. The equation that describes the curve is

Which one of the following is the general solution of the first order differential equation $${{dy} \over {dx}} = {\left( {x + y - 1} \right)^2}$$ , where $$x,$$ $$y$$ are real ?

The particular solution of the initial value problem given below is $$\,\,{{{d^2}y} \over {d{x^2}}} + 12{{dy} \over {dx}} + 36y = 0\,\,$$ with $$\,y\left( 0 \right) = 3\,\,$$ and $$\,\,{\left. {{{dy} \over {dx}}} \right|_{x = 0}} = - 36\,\,$$

Consider the differential equation $${{{d^2}x\left( t \right)} \over {d{t^2}}} + 3{{dx\left( t \right)} \over {dt}} + 2x\left( t \right) = 0$$
Given $$x(0) = 20$$ & $$\,x\left( 1 \right) = {{10} \over e},$$ where $$e=2.718,$$

The value of $$x(2)$$ is